hello i have a question about my buffer an pH LAB SO i am suppose to find the in

Question

hello i have a question about my buffer an pH LAB

SO i am suppose to find the inital moles and acetate acd ratio using ice

pH

.20M HC2H3O2 2.97 used 25mL of it and diluted with DI water 50mL

.20M NaC2H3O2 6.30 same

Combined two 4.85

combination is below

pH

trial vial #

1 4.84

2 4.47

3 4.90

4 3.2 0

5 11.67

6 4.4

7 5.43

8 5.14

9 4.32

10 4.70

concentration i got 0.20M *25mL/50m;/L=.01M

so i was gven M1V1/V2

how do i find the intial moles and ratio of acetate using ice. please show all steps.for both problems someone did a great job solving for it but did not show some steps which confused me. fo

10.Manual pdf Xe Masteringchemistry Ais xG how to take a screensho x C Chegg Study IGuided S G What is the con X 102 M ants-chem t/Classes/102/manual/102 Manua 5 41 131 102 Manual.pdf 12. Use clean 10 mL graduated cylinders and follow the chart to carefully measure the volume of each reagent indicated into separate, well-drained shell vials. Shell combined 20-M O.200-M o 10-M Vial sodium Number Solution acetic acid acetate 0.10-M NaOH 3.0 mL 2.0 mL 2.0 mL 3.0 mL 4 3.0 mL 3.0 mL 3.0 mL 2.0 mL 4.0 mL 2.0 mL 3.0 mL 3.0 mL 4.0 mL 2.0 mL 3.0 mL 1o 2.0 mL 4.0 mL 11 12 Cover each shell vial with parafilm Miv the contents well hv inversion then meas 91% 1:30 PM

Explanation / Answer

There are three ways of making buffer-

1) By mixing the acid and its conjugate base in the required ratio as per the calculations using Henderson-hasselbach equation

2) By taking acid and partly neutralizing it by adding strong base like NaOH to the required point to get a combination of acid and its conjugate base

3)By taking salt of the acid (sodium acetate) and partly neutralizing it by adding strong acid like HCl to the required point to get a combination of acid and its conjugate base

As you have combined acid and its conjugate base ,you have implemented the first method.

i m doing calculations for trial 1

seeing your table i assume that you want to make 4.00 ml of combined solution or total solution.You can change this and repeat calculations (if any changes are there)

Henderson-hasselbach equation can be used to calculate the ratio of moles of HC2H3O2 and its conjugate base ,NaC2H3O2 .

pH=pka +log [conjugate base]/[acid]

or, pH=pka +log [NaC2H3O2]/[HC2H3O2]

pka for acetic acid =4.75

trial 1 ,pH=4.84

pH=pka +log [NaC2H3O2]/[HC2H3O2]

or, 4.84=4.75 +log [NaC2H3O2]/[HC2H3O2]

[NaC2H3O2]/[HC2H3O2]=1.230 [molar ratio]

concentration using M1V1/V2= .20M* 25mL/50mL =0.1M

As both NaC2H3O2 and HC2H3O2 used had the same molarity,so, [NaC2H3O2]/[HC2H3O2]=1.230=0.1M*V1/0.1M*V2=V1/V2

So,V1/V2=1.23 or, V1=1.23*V2

if total volume =4.0 ml=V1+V2

then 4.0 ml=1.23V2+V2=2.23V2

V2=4.0/2.23=1.8 ml

V1=4ml-1.8ml=2.2 ml

So V1=volume of NaC2H3O2 =2.2ml

V2=volume of HC2H3O2=1.8ml

So initial mole of NaC2H3O2=0.1M*2.2ml=0.1 mol/L *2.2*10^-3L=2.2*10^-4 moles

initial moles of HC2H3O2=0.1M*1.8ml=0.1 mol/L *1.8*10^-3L=1.8*10^-4 moles

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